Answer
$x=\left\{ \dfrac{-5-\sqrt{41}}{8},\dfrac{-5+\sqrt{41}}{8}
\right\}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the solutions of the given equation, $
(2x-1)^2=x+2
,$ express first in the form $ax^2+bx+c=0.$ Then use the Quadratic Formula.
$\bf{\text{Solution Details:}}$
Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
(2x)^2-2(2x)(1)+(1)^2=x+2
\\\\
4x^2-4x+1=x+2
.\end{array}
Using the properties of equality, in the form $ax^2+bx+c=0,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
4x^2+(-4x-x)+(1-2)=0
\\\\
4x^2-5x-1=0
.\end{array}
The quadratic equation above has $a=
4
, b=
-5
, c=
-1
.$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then
\begin{array}{l}\require{cancel}
x=\dfrac{-5\pm\sqrt{(-5)^2-4(4)(-1)}}{2(4)}
\\\\
x=\dfrac{-5\pm\sqrt{25+16}}{8}
\\\\
x=\dfrac{-5\pm\sqrt{41}}{8}
.\end{array}
The solutions are
\begin{array}{l}\require{cancel}
x=\dfrac{-5-\sqrt{41}}{8}
\\\\\text{OR}\\\\
x=\dfrac{-5+\sqrt{41}}{8}
.\end{array}
Hence, $
x=\left\{ \dfrac{-5-\sqrt{41}}{8},\dfrac{-5+\sqrt{41}}{8}
\right\}
.$
